Question

Pls answer me.
physics.​

Answer 1

Answer:

Explanation: In a velocity time graph, the slope of graph tell us about the acceleration. If the slope is high then the acceleration is positive, if the solve is low then the acceleration is negative and if the slope is parallel to time axis then there is no acceleration taking place!

Explanation: In a velocity time graph if we have to find out the distance or displacement then we have to find out the area under the curve!

• We know that distance and displacement are same in the case of graph.

Required solution:

a) The distance and displacement of the particle from 0 to 15 seconds is 100 m and 100 m.

Firstly let us find out the area of triangle

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \: B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times (10-0) \times (15-0) \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 10 \times 15 \\ \\ :\implies \sf Area \: of \: \triangle \: = 1 \times 5 \times 15 \\ \\ :\implies \sf Area \: of \: \triangle \: = 75 \: m^2$

Now let's find out the area of square

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: square \: = a \times a \\ \\ :\implies \sf Area \: of \: square \: = (15-10) \times (5-0) \\ \\ :\implies \sf Area \: of \: square \: = 5 \times 5 \\ \\ :\implies \sf Area \: of \: square \: = 25 \: m^2$

Now finding total distance or displacement

$:\implies \sf s \: = Areas \\ \\ :\implies \sf s \: = 75 + 25 \\ \\ :\implies \sf s \: = 100 \: m \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

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b) The distance and displacement of the particle between 10 s to 20 s is 50 m and 50 m.

Finding area of triangle

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \: B \times H \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times (20-15) \times (10-0) \\ \\ :\implies \sf Area \: of \: \triangle \: = \dfrac{1}{2} \times 5 \times 10 \\ \\ :\implies \sf Area \: of \: \triangle \: = 1 \times 5 \times 5 \\ \\ :\implies \sf Area \: of \: \triangle \: = 25 \: m^2$

Now area of square

$:\implies \sf s \: = Area \: under \: curve \\ \\ :\implies \sf Distance \: = Area \: under \: curve \\ \\ :\implies \sf Area \: of \: square \: = a \times a \\ \\ :\implies \sf Area \: of \: square \: = (15-10) \times (5-0) \\ \\ :\implies \sf Area \: of \: square \: = 5 \times 5 \\ \\ :\implies \sf Area \: of \: square \: = 25 \: m^2$

Now finding total distance or displacement

$:\implies \sf s \: = Areas \\ \\ :\implies \sf s \: = 25 + 25 \\ \\ :\implies \sf s \: = 50 \: m \\ \\ {\pmb{\sf{Henceforth, \: solved!}}}$

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c) In which time interval the particle is speeding up?

→ In time interval of 0 to 10 seconds the particle is increasing its speed.

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d) In which time interval the particle is moving with zero acceleration?

→ In the time interval 10 to 15 seconds the particle is moving with zero acceleration and we also know that if the velocity is constant then acceleration must be zero.

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e) In which time interval the particle is moving with negative acceleration?

→ In the time interval 15 to 20 seconds it is moving with negative acceleration.